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3 years ago

How can we prevent chemical hazards in labotary

Engineering

1 answer:

Maurinko [17]3 years ago

4 0

Answer:

We can prevent it by:

a) By wearing GOOGLES.

b) By wearing our Lab coat.

c) Fire extinguisher should always be present in the lab.

d) Hand Gloves must be worn.

e) No playing in the lab.

f) No touching of things/equipment's e.g bottles, in the lab.

g) No eating/snacking in the lab.

h) Always pay attention, no gisting.

i) Adult/qualified person must be present in the lab with pupils/students.

Explanation:

Hope it helps.

You might be interested in

Saturated water vapor undergoes a throttling process from 1bar to a 0.35bar. What is the change in temperature for this process?

mamaluj [8]

Answer:

-25.63°C.

Explanation:

We know that throttling is a constant enthalpy process

$h_1=h_2$

From steal table

We know that if we know only one property in side the dome then we will find the other property by using steam property table.

Temperature at saturation pressure 1 bar is 99.63°C and Temperature at saturation pressure 0.35 bar is about 74°C .

So from above we can say that change in temperature is -25.63°C.

But there is no any option for that .

4 0

3 years ago

A closed, rigid tank is lled with a gas modeled as an ideal gas, initially at 27°C and a gage pressure of 300 kPa. The gas is he

sergejj [24]

Answer:

T₂ =93.77 °C

Explanation:

Initial temperature ,T₁ =27°C= 273 +27 = 300 K

We know that

Absolute pressure = Gauge pressure + Atmospheric pressure

Initial pressure ,P₁ = 300+1=301 kPa

Final pressure ,P₂= 367+1 = 368 kPa

Lets take temperature=T₂

We know that ,If the volume of the gas is constant ,then we can say that

$\dfrac{P_2}{P_1}=\dfrac{T_2}{T_1}$

${T_2}=T_1\times \dfrac{P_2}{P_1}$

Now by putting the values in the above equation we get

${T_2}=300\times \dfrac{368}{301}\ K$

The temperature in °C

T₂ = 366.77 - 273 °C

T₂ =93.77 °C

8 0

3 years ago

An air-conditioning system operating on the reversed Carnot cycle is required to transfer heat from a house at a rate of 755 kJ/

Lyrx [107]

Answer:

There is 0.466 KW required to operate this air-conditioning system

Explanation:

Step 1: Data given

Heat transfer rate of the house = Ql = 755 kJ/min

House temperature = Th = 24°C = 24 +273 = 297 Kelvin

Outdoor temperature = To = 35 °C = 35 + 273 = 308 Kelvin

Step 2: Calculate the coefficient of performance o reversed carnot air-conditioner working between the specified temperature limits.

COPr,c = 1 / ((To/Th) - 1)

COPr,c = 1 /(( 308/297) - 1)

COPr,c = 1/ 0.037

COPr,c = 27

Step 3: The power input cna be given as followed:

Wnet,in = Ql / COPr,max

Wnet, in = 755 / 27

Wnet,in = 27.963 kJ/min

Win = 27.963 * 1 KW/60kJ/min = 0.466 KW

There is 0.466 KW required to operate this air-conditioning system

3 0

3 years ago

A large retirement village has a total retail employment of 120. All 1600 of the households in this village consist of two nonwo

polet [3.4K]

Answer:

See explaination

Explanation:

Please kindly check attachment for the step by step solution of the given problem.

5 0

4 years ago

Un mol de gas ideal realiza un trabajo de 3000 J sobre su entorno, cuando se expande de manera isotermica a una temperatura de 5

Shkiper50 [21]

Answer:

74,4 litros

Explanation:

Dado que

W = nRT ln (Vf / Vi)

W = 3000J

R = 8,314 JK-1mol-1

T = 58 + 273 = 331 K

Vf = desconocido

Vi = 25 L

W / nRT = ln (Vf / Vi)

W / nRT = 2.303 log (Vf / Vi)

W / nRT * 1 / 2.303 = log (Vf / Vi)

Vf / Vi = Antilog (W / nRT * 1 / 2.303)

Vf = Antilog (W / nRT * 1 / 2.303) * Vi

Vf = Antilog (3000/1 * 8,314 * 331 * 1 / 2,303) * 25

Vf = 74,4 litros

3 0

3 years ago